The graph of the left side of the equation is shown below:
The graph of the right side of the equation is shown below:
Comparing them we notice that they are the same graph, therefore the equation is an identiy.
To prove the idenity we need to remember that:
[tex]\begin{gathered} \csc \theta=\frac{1}{\sin \theta} \\ \cot \theta=\frac{\cos \theta}{\sin \theta} \end{gathered}[/tex]
also, we need to remember that:
[tex]1-\cos ^2\theta=\sin ^2\theta[/tex]
Let's use this properties on the left side of the equation:
[tex]\begin{gathered} (\csc \theta+\cot \theta)(1-\cos \theta)=(\frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta})(1-\cos \theta) \\ =(\frac{1+\cos \theta}{\sin ^{}\theta})(1-\cos \theta) \\ =\frac{(1+\cos \theta)(1-\cos \theta)}{\sin \theta} \\ =\frac{1-\cos ^2\theta}{\sin ^{}\theta} \\ =\frac{\sin ^2\theta}{\sin \theta} \\ =\sin \theta \end{gathered}[/tex]
Therefore:
[tex](\csc \theta+\cot \theta)(1-\cos \theta)=\sin \theta[/tex]
and we proved what we noticed on the graphs.
